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Question Number 86374 by mathmax by abdo last updated on 28/Mar/20

$$calculatebycomplexmethod{\int }_1^{+\mathrm{\infty }}\frac{dx}{1+x^2}$$

Commented by mathmax by abdo last updated on 28/Mar/20

$${\int }_1^{+\mathrm{\infty }}\frac{dx}{x^2+1}={\int }_1^{+\mathrm{\infty }}\frac{dx}{(x-i)(x+i)}=\frac{1}{2i}{\int }_1^{+\mathrm{\infty }}(\frac{1}{x-i}-\frac{1}{x+i})dx$$$$=\frac{1}{2i}{\left[ln\left(\frac{x-i}{x+i}\right)\right]}_1^{+\mathrm{\infty }}=\frac{1}{2i}(-ln\left(\frac{1-i}{1+i}\right))=\frac{1}{2i}ln\left(\frac{1+i}{1-i}\right)wehave$$$$\frac{1+i}{1-i}=\frac{\sqrt{2}{e}^{\frac{i\pi }{4}}}{\sqrt{2}{e}^{-\frac{i\pi }{4}}}=e^{\frac{i\pi }{2}}=\Rightarrow ln\left(\frac{1+i}{1-i}\right)=\frac{i\pi }{2}\Rightarrow$$$${\int }_1^{+\mathrm{\infty }}\frac{dx}{x^2+1}=\frac{1}{2i}\times \frac{i\pi }{2}=\frac{\pi }{4}$$

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